给定一个矩阵

[
  [1 ,2 ,3 ,4 ,5],
  [16,17,24,23,6],
  [15,18,25,22,7],
  [14,19,20,21,8],
  [13,12,11,10,9]
]

返回 25

思路：记忆化搜索 + dp

　　设Lics(num)表示以num开头的最长上升子连续序列的长度， 则Lics(A[x][y]) = max(Lics(A[x-1][y]), Lics(A[x][y-1]), Lics(x+1,y), Lics(x, y+1))+1;

#include<iostream> 
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<deque>
#include<map> 
using namespace std;
 
class Solution {
public:
    /**
     * @param A an integer matrix
     * @return  an integer
     */
    int n, m, maxL;
    int lics[1000][1000];
    int vis[1000][1000];
    int dir[4][2] = {{0, 1}, {1, 0}, {0,-1}, {-1,0}};
    int dfs(vector<vector<int>>& A, int x, int y){
        int maxLics = 0;
        vis[x][y] = 1;
        for(int i=0; i<4; ++i){
            int xx = x+dir[i][0];
            int yy = y+dir[i][1];
            if(xx<0 || yy<0 || xx>=n || yy>=m) continue;
            if(A[x][y] >= A[xx][yy]) continue;
            if(!vis[xx][yy])
                maxLics = max(maxLics, dfs(A, xx, yy));
            else
                maxLics = max(maxLics, lics[xx][yy]);
        }
        lics[x][y] = maxLics+1;
        if(maxL < lics[x][y]) maxL = lics[x][y];
        return lics[x][y];
    } 
    
    int longestIncreasingContinuousSubsequenceII(vector<vector<int>>& A) {
        n = A.size();
        if(n == 0) return 0;
        m = A[0].size();
        memset(lics, 0, sizeof(lics));
        memset(vis, 0, sizeof(vis));
        maxL = 0;
        for(int i=0; i<n; ++i)
            for(int j=0; j<m; ++j)
                if(!vis[i][j])
                    dfs(A, i, j);
        return maxL;
    }
};
/*
2 3 4 5
17 24 23 6
18 25 22 7
19 20 21 8
12 11 10 9
*/